Match each ether in Column I with the products obtained when one mole is cleaved with excess HI in the cold or on gentle warming, taking the usual SN1 or SN2 rule. Column I: (P) CH3-O-CH(CH3)2; (Q) (CH3)3C-O-CH3; (R) C6H5-O-CH2CH3; (S) tetrahydrofuran. Column II: (1) CH3I and (CH3)2CHOH; (2) (CH3)3C-I and CH3OH; (3) C6H5OH and CH3CH2I; (4) I-(CH2)4-I (with excess HI).

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Match each ether in Column I with the products obtained when one mole is cleaved with excess HI in the cold or on gentle warming, taking the usual SN1 or SN2 rule. Column I: (P) $CH X_{3} - O - CH (CH X_{3}) X_{2}$ ; (Q) $(CH X_{3}) X_{3} C - O - CH X_{3}$ ; (R) $C X_{6} H X_{5} - O - CH X_{2} CH X_{3}$ ; (S) tetrahydrofuran. Column II: (1) $CH X_{3} I$ and $(CH X_{3}) X_{2} CHOH$ ; (2) $(CH X_{3}) X_{3} C - I$ and $CH X_{3} OH$ ; (3) $C X_{6} H X_{5} OH$ and $CH X_{3} CH X_{2} I$ ; (4) $I - (CH X_{2}) X_{4} - I$ (with excess HI).

AP-1, Q-2, R-3, S-4
BP-2, Q-1, R-3, S-4
CP-1, Q-2, R-4, S-3
DP-1, Q-3, R-2, S-4

Answer: A

P is a methyl secondary ether and iodide attacks the unhindered methyl carbon (SN2), giving

CH X_{3} I

and isopropanol. Q cleaves by SN1 at the tertiary carbon. R cleaves at the ethyl carbon because the aryl carbon cannot be attacked. THF has two primary carbons, and with excess HI both are attacked and the diol-type intermediate becomes 1,4-diiodobutane. The match is P-1, Q-2, R-3, S-4.

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Answer: A

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