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Question of 135

Q.(a) Complete the following reaction:
CH3-CH2-CH2-CH2-O-CH2-CH3 + HI →

(b) Explain why propanol has higher boiling point than that of the hydrocarbon butane.
Chhattisgarh CgbseCGBSE Intermediate Board 2025Subjective· 3mImportance★★★★★
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(a) HI cleaves the unsymmetrical ether by SN2 attack of I⁻ on the smaller (ethyl) alkyl group, giving butan-1-ol and ethyl iodide. (b) Propan-1-ol boils higher than butane because it can hydrogen-bond while butane cannot.

(a) CH3-CH2-CH2-CH2-O-CH2-CH3 (an unsymmetrical ether, 1-ethoxybutane) + HI:

Ethers are cleaved by strong acids like HI via protonation of the ether oxygen followed by nucleophilic attack of I⁻ on one of the alkyl carbons.

Both alkyl groups here (butyl and ethyl) are primary, so cleavage proceeds by the SN2 mechanism, and the nucleophile (I⁻) attacks the less hindered (smaller) alkyl group preferentially — i.e., the ethyl carbon rather than the bulkier butyl carbon.

With one equivalent (limited/cold) HI:

CH3CH2CH2CH2-O-CH2CH3 + HI → CH3CH2CH2CH2-OH (butan-1-ol) + CH3CH2-I (iodoethane)

(With excess/hot HI, the alcohol formed would itself be further converted to the corresponding alkyl iodide: CH3CH2CH2CH2OH + HI → CH3CH2CH2CH2I + H2O.)

(b) Why propanol has a higher boiling point than butane: …

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