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Hydrolysis of neopentyl bromide under $SN_1$ process results into


$ (a)\;(i)\; and\;(ii) \\ (b) \; (ii)\;and \;(iii) \\ (c) \;(ii)\;and \;(iv) \\ (d)\;(iii)\;and (iv) $
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$H_2O$ is neutral nucleophile and $3^{\circ}$ alkyl halide with neutral nucleophile undergoes elimination as well as substitution.
Hence a is the correct answer.
answered Feb 28, 2014 by meena.p

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