Skip to content
Question of 147

Q.Complete the following reactions:

(i) CH3−CH2−Br+KCN⟶CH_3-CH_2-Br + KCN \longrightarrow ? (1 mark);
(ii) chlorobenzene (benzene ring with a ClCl substituent) →Anh AlCl3CH3−CO−Cl\xrightarrow[\text{Anh } AlCl_3]{CH_3-CO-Cl} ? (Friedel–Crafts acylation) (1 mark) OR In the following pair of halogen compounds, which compound will react faster by SN1S_N1 mechanism? Why? (1+1 marks) — (CH3)3C−Cl(CH_3)_3C-Cl (tert-butyl chloride: a central carbon bonded to three CH3CH_3 groups and one ClCl) Or (CH3)2CH−Cl(CH_3)_2CH-Cl (isopropyl-type chloride: a central carbon bonded to two CH3CH_3 groups, one HH, and one ClCl)
Meghalaya MboseMBOSE Meghalaya Intermediate Board 2022Subjective· 2mImportance★★★★★
0% · 0/147 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

KCNKCN substitutes bromide via its nucleophilic carbon to give an alkyl nitrile, and anhydrous AlCl3AlCl_3-catalysed Friedel–Crafts acylation on chlorobenzene installs an acetyl group mainly at the para position; in the alternative question, the more stable tertiary carbocation makes tert-butyl chloride the faster SN1S_N1 substrate.

(i) CH3CH2Br+KCNCH_3CH_2Br + KCN

CH3−CH2−Br+KCN→CH3−CH2−C≡N+KBrCH_3-CH_2-Br + KCN \xrightarrow{} CH_3-CH_2-C\equiv N + KBr

KCNKCN is an ionic compound in which the cyanide ion is an ambident nucleophile (can attack through either CC or NN). Because the carbon atom of CN−CN^- is the site of higher electron density/greater nucleophilicity in the free (ionic) cyanide ion, nucleophilic substitution occurs mainly through carbon, giving the alkyl cyanide (nitrile), propanenitrile (CH3CH2CNCH_3CH_2CN, also called ethyl cyanide), as the major product — rather than the isocyanide (isonitrile) that would form if NN attacked. (By contrast, the more covalent AgCNAgCN favours attack through NN, giving isocyanides, R−NCR-NC.)

(ii) Chlorobenzene + CH3COClCH_3COCl / anhydrous AlCl3AlCl_3

This is a Friedel–Crafts acylation: anhydrous AlCl3AlCl_3 (a Lewis acid) polarises CH3−CO−ClCH_3-CO-Cl to generate the electrophilic acylium ion CH3−CO+CH_3-CO^+, which attacks the benzene ring of chlorobenzene.

C6H5Cl+CH3COCl→Δanhyd. AlCl3p-Cl-C6H4-CO-CH3 (+ o-isomer, minor)+HCl\text{C}_6H_5Cl + CH_3COCl \xrightarrow[\Delta]{\text{anhyd. } AlCl_3} p\text{-Cl-C}_6H_4\text{-CO-CH}_3\ (+\ o\text{-isomer, minor}) + HCl

The −Cl-Cl substituent is a weak deactivator but an o,po,p-director (its lone pair can be donated into the ring by resonance even though its −I-I effect withdraws electron density overall), so the incoming acetyl group is directed to the ortho and para positions. Because the acetyl (bulky, −COCH3-COCH_3) group and AlCl3AlCl_3-complexation create significant steric hindrance at the position adjacent to ClCl, the para product, 1-(4-chlorophenyl)ethan-1-one (p-chloroacetophenone), dominates over the ortho isomer.

Alternative (Or): SN1S_N1 reactivity of (CH3)3CCl(CH_3)_3CCl vs (CH3)2CHCl(CH_3)_2CHCl

SN1S_N1 rate depends on carbocation stability; the tertiary carbocation from tert-butyl chloride is far more stable than the secondary carbocation from isopropyl chloride, so tert-butyl chloride reacts faster.

…

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.