by Adam Le Gresley, PhD

Questions about the lecture
My Notes
  • Required.
Save Cancel
    Learning Material 2
    • PDF
      Slides 04 Chemistry Advanced Le Gresley.pdf
    • PDF
      Download Lecture Overview
    Report mistake
    Okay, now, I want to bring you on to the lecture that deals with Haloalkanes. Over the course of the Module III, you’ll see that we’re going to be covering a number of different so called “functional groups”. These are groups which are important in terms of not only synthetic pathways, but also in terms of biological application, medical application that results from that. The general formula for haloalkanes is R-X, where X is one of the halogens either fluorine, chlorine, bromine or iodine. But, before we move on to how they react and with what, let’s just have a quick look at their nomenclature. As you can see, at the bottom of the board, we’ve got two different types of haloalkanes. We have 2-bromobutane and we have also 3-bromo-4-ethylheptane. And I want to briefly talk to you a little bit about priority, when it comes to assigning them a nomenclature. This is important not just for the compounds that you produce, but also when you’re using starting material that is labelled accordingly. Typically speaking, when you’re looking at something which has a large atomic mass in comparison to carbon, it takes priority. So, if you look at Bromine, for example, with a mass of over 80 grams per mole, it takes priority. And the numbering goes from the closest terminal carbon to that substituent; the closest being the 1-carbon away, which we’ve denoted 1 in the case of 2-bromobutane. We then count along 1, 2, 3 and 4. And so, we can therefore say, bromine, where it’s positioned is 2. And therefore, as a halogen, the halogen component appears as a prefix to the alkane chain itself: Halo-ane. If we look at 3-bromo-4-ethylheptane, we’ll see the same thing. The nearest terminal carbon to where the bromine is substituted is...

    About the Lecture

    The lecture Haloalkanes by Adam Le Gresley, PhD is from the course Organic Chemistry. It contains the following chapters:

    • Haloalkanes
    • Nucleophilic substitution
    • Reaction parameters
    • SN 1 Reaction
    • Summary SN 1 vs. SN 2
    • Eliminiation reactions

    Included Quiz Questions

    1. 2-Fluoro-3-methylpentane
    2. 3-Methyl-4-fluoropentane
    3. 2-Fluorohexane
    4. 2-Fluoro-3-ethyl-3-methylpropane
    5. 2-Fluoro-3-ethylbutane
    1. …carbon chain length of 2-Chlorobutane.
    2. …electronegativity of chlorine.
    3. …strength of sigma bond between carbon and chlorine.
    4. …stability of the generated chloride.
    5. …electronegativity of halogen.
    1. …Pauling electronegativity scale.
    2. …Allen electronegativity scale.
    3. …Mulliken-Jaffe electronegativity scale.
    4. …Sanderson electronegativity scale.
    5. …Allred Rochow electronegativity scale.
    1. ...F > Cl > Br > I.
    2. …I > Br > Cl > F.
    3. …Cl > Br > I > F.
    4. …Br > I > Cl > F.
    5. …F > I > Br > Cl.
    1. …high electronegativity of F.
    2. …high electronegativity of C.
    3. …small atomic size of C.
    4. …presence of H atoms in 2-fluorobutane.
    5. …electron pulling effect of H atom.
    1. …has an electron withdrawing inductive effect.
    2. …does not affect the distribution of electron density on the haloalkane molecule.
    3. …repels the electron density toward other atoms.
    4. …does not participate in a chemical reaction.
    5. …has an electron releasing inductive effect.
    1. …C-F > C-Cl > C-Br > C-I.
    2. …C-I > C-Br > C-Cl > C- F.
    3. …C-Br > C-I > C-F > C-Cl.
    4. …C-F < C-Cl < C-I < C-Br.
    5. …C-I < C-Br < C-F < C-Cl.
    1. …weaker bond strength between C and I.
    2. …high electronegativity of F.
    3. …presence of I at the C-2 position.
    4. …presence of F at the C-2 position.
    5. …absence of a substituent alkyl group near the F.
    1. …HF < HCl < HBr < HI.
    2. …HF < HCl < HI < HBr.
    3. …HF > HCl> HBr > HI.
    4. …HF > HBr > HCl > HI.
    5. … HBr < HCl < HF < HI.
    1. …the strength of H-X bond.
    2. …the temperature conditions.
    3. …the presence of metal.
    4. …the atomic mass of hydrogen atom.
    5. …the atomic radius of the hydrogen atom.
    1. CH3CHBrCH3
    2. CH3CH2CH2Br
    3. CH3CHBrCH2Br
    4. No Reaction
    5. BrCH2CH2CH2Br
    1. …the electron deficient carbon of a haloalkane.
    2. …the electron dense carbon of a haloalkane.
    3. …the electron dense halogen of a haloalkane.
    4. …the electron deficient halogen of a haloalkane.
    5. …the hydrogen atom of the terminal carbon on a haloalkane.
    1. …leaves as a halide ion.
    2. … leaves as a halogen atom.
    3. …leaves as a hydrohalic acid.
    4. …switches its position from the terminal to the central carbon atom.
    5. …displaces the hydrogen from haloalkane.
    1. NaOH.
    2. H2O.
    3. CH3OH.
    4. H2S.
    5. CH3SH.
    1. …stererospecific.
    2. …unimolecular.
    3. …only substrate concentration dependent.
    4. …only nucleophile concentration dependent.
    5. …controlled by the leaving halide.
    1. …the reaction rate depends on both nucleophile and substrate concentrations.
    2. …OH- ions are produced in excess.
    3. …Na+ ions extract the bromide from the substrate.
    4. …bromide stimulates the dissociation of NaOH into ions.
    5. …NaOH acts as a catalyst.
    1. The neutral nucleophile species are more nucleophilic than negatively charged nucleophiles.
    2. A nucleophile has a negative charge or partial negative charge or an electron pair or pi electron available.
    3. A nucleophile donates an electron pair to an electrophile to form a covalent bond.
    4. RO-, R2N-, HS-, RS- and Cl- are negatively charged nucleophiles.
    5. H2O, NH3, RNH2 and R3N are neutral nucleophiles.
    1. BF3
    2. CH3O-
    3. NH3
    4. CH3OH
    5. CN-
    1. I-> Br- > Cl- > F-
    2. F- > Cl- > Br- > I-
    3. Cl- > Br- > I- > F-
    4. F- > Cl- > I- > Br-
    5. F- > I- > Br- > Cl-
    1. Nucleophilicity trend for halogens is I < Cl < Br < F.
    2. Water is weaker nucleophile than alcohol.
    3. CN- exhibits more nucleophilicity than NH3.
    4. Carbon has more nucleophilic character than oxygen.
    5. Basicity trend of halogens is F > Cl > Br > I.
    1. C- > N- > O- > F-
    2. F- > O- > N-> C-
    3. O- > N- > C- > F-
    4. N- > C- > F- > O-
    5. C- > F- > N- > O-
    1. Dimethylformamide (DMF)
    2. Formic acid
    3. Hydrogen fluoride
    4. Ammonia
    5. n-Butanol
    1. Alcohol
    2. Dimethylsulphoxide (DMSO)
    3. Dimethylformamide (DMF)
    4. Tetrahydrofuran (THF)
    5. Ethylacetate
    1. …they solvate the cations, thus leaving the anions more nucleophilic.
    2. …they solvate the anions, thus leaving the cations for nucleophilic.
    3. …they act as a catalyst during the SN2 reaction.
    4. …they solvate the cations, thus leaving the anions more electrophilic.
    5. …the solvate the anions, thus leaving the cations more electrophilic.
    1. CH3- > CH3-CH2- > (CH3)2-CH- > (CH3)3-C-.
    2. (CH3)3-C- > CH3-CH2- > (CH3)2-CH- > CH3-.
    3. (CH3)3-C- > (CH3)2-CH- > CH3-CH2- > CH3- .
    4. CH3- > CH3-CH2- > (CH3)3-C- > (CH3)2-CH-.
    5. (CH3)3-C- > (CH3)2-CH- > CH3- > CH3-CH2-.
    1. Due to sterically less congested structure
    2. Because of the high degree of deprotonation of bromomethane
    3. Because of the high degree of deprotonation of 2-bromopropan
    4. Due to the nucleophilic attack from the front side of the haloalkane
    5. Due to the presence of negative charge on carbon in bromomethane
    1. Fluorine is a good leaving group with weak basicity.
    2. The haloalkanes with strong acidity character produce weaker conjugate bases.
    3. The conjugate base of a strong acid is a good leaving group.
    4. The stability of leaving group affects the reactivity of haloalkane.
    5. The Br- produced during the SN2 reaction between 2-bromobutane and NaOH is also technically a nucleophile.
    1. I-, Br-, Cl-, RCOO-, tosylates and mesylates are not good leaving groups during SN2 reactions.
    2. F-, HO-, CH3O-, NH2-, H- and C2H5- are poor leaving groups during the SN2 reaction.
    3. Tertiary alkyl halides do not give the SN2 type reactions
    4. The polar protic solvents are avoided during SN2 type reactions due to their anion solvation tendencies.
    5. The ethanol decreases the SN2 type reaction rate dramatically by solvating the HO- and making it unavailable to attack the alkyl halide.
    1. …an enantiomer into the racemic mixture.
    2. …an S-enantiomer into an R-enantiomer.
    3. …an R enantiomer into an S- enantiomer.
    4. …a primary alkyl halide into a mixture of secondary and tertiary alkyl halides.
    5. …a tertiary alkyl halide into a secondary alkyl halide.
    1. The reaction rate of SN1 type reaction depends upon the concentrations of both alkyl halide and the attacking nucleophile.
    2. The bromide ion of 3-bromo-3-propyloctane leaves at first place during the SN1 reaction; then the incoming nucleophile (OH-) attacks the generated carbocation.
    3. The bromide ion losing-step (unimolecular) is the rate-determining step in above reaction.
    4. The product consists of a mixture of S and R enantiomers of 3-Propyloctan-3-ol.
    5. An intermediate carbocation is generated during SN1 type reaction.
    1. The carbocation intermediate contains negative charge on the carbon atom due to the presence of a lone pair of electrons.
    2. The carbocation intermediate has an sp2 hybridized carbon with the trigonal planar arrangement.
    3. Due to the trigonal planar arrangement of carbocation intermediate, the incoming nucleophile can attack the either side of the carbocation.
    4. The carbocation formation is not favored energetically, hence this is a slow and rate determining step.
    5. The carbocation intermediate has a positively charged carbon atom.
    1. …ion pair effect.
    2. …trigonal planar arrangement of carbon atom of carbocation intermediate.
    3. …high concentration of attacking nucleophile.
    4. …unfavorable temperature conditions of the reaction mixture.
    5. …presence of an impurity or heavy metal in the reaction mixture.
    1. The rate of SN1 reaction depends upon the nucleophilicity of the nucleophile.
    2. The rate of SN1 reaction depends upon the stability of the carbocation intermediate.
    3. With an increase in the alkyl substitution of the positively charged carbon of carbonation, the probability of SN1 reaction increases.
    4. The polar protic solvents usually favor SN1 type reactions in tertiary haloalkanes.
    5. A strong leaving group in haloalkane favors the SN1 type reaction.
    1. …tertiary > secondary > primary > methyl.
    2. …secondary > primary > methyl > tertiary.
    3. …methyl > primary > secondary > tertiary.
    4. …primary > secondary > tertiary > methyl.
    5. …secondary > tertiary > primary > methyl.
    1. Ethanol
    2. Toluene
    3. Benzene
    4. DMSO
    5. THF
    1. The fluoroalkanes give faster SN2 and SN1 type reactions with NaOH in the presence of a non-polar solvent.
    2. The polar protic solvents favor SN1 type reactions because of they solvate the nucleophile by forming a solvent shell around it.
    3. 2-Chloro-2-methylpentane undergoes an SN1 type reaction more readily than 1-Chloro-pentane.
    4. The electron-releasing inductive effect of alkyl groups in a highly substituted haloalkane impacts the stability of carbocation.
    5. 2-Iodo-2-ethyloctane gives a faster SN1 reaction than 2-chloro-2-ethyloctane. as I- is a better leaving group than Cl-.
    1. Trans or Entgegen (E)
    2. Cis or Zusammen (Z)
    3. Racemisation
    4. No preference
    5. Mixture of Z and E
    1. …beta (β).
    2. …alpha (α).
    3. …gamma (γ).
    4. …epsilon (ε).
    5. …zeta (ζ).
    1. Propene
    2. Propan-2-ol
    3. Propan-1-ol
    4. Propyne
    5. 1-Bromopropane
    1. …concentrations of both 2-bromobutane and NaOH.
    2. …concentration of 2-bromobutane.
    3. …concentration of NaOH.
    4. …concentration of the solvent.
    5. …concentrations of alcohol.
    1. …formation of a pi double bond between two carbon atoms and releases of halogen as a halide ion.
    2. … formation of a triple bond between two carbon atoms and releases of halogen as a halide ion.
    3. … formation of a triple bond between two carbon atoms and releases of a halogen atom.
    4. … formation of a double bond between two carbon atoms and releases of a halogen atom.
    5. …formation of an alkane with the release of halogen as a halide ion.
    1. …the most stable and highly substituted internal alkene.
    2. …the most stable and highly substituted terminal alkene.
    3. …the most unstable and highly substituted internal alkene.
    4. …the most unstable and highly substituted terminal alkene.
    5. …the most stable and highly substituted terminal alkyne.
    1. …elimination of halide ion takes place in the absence of a base.
    2. …elimination of halide ion depends upon the concentrations of both haloalkane and base.
    3. …elimination of halide ion depends upon the base concentration only.
    4. …formation of terminal alkene dependent upon the temperature of the reaction mixture.
    5. …alcoholic KOH determines the stability of generated carbocation.

    Author of lecture Haloalkanes

     Adam Le Gresley, PhD

    Adam Le Gresley, PhD

    Customer reviews

    5,0 of 5 stars
    5 Stars
    4 Stars
    3 Stars
    2 Stars
    1  Star