Which of the following alkyl halides is preferred for the synthesis of ether by Williamson synthesis?

n-Propyl bromide

The correct answer is option 1. n-Propyl bromide.

In Williamson synthesis, for optimal results, the alkyl halide used should be a:

Primary alkyl halide: Primary alkyl halides (R-X, where R is a primary alkyl group) are preferred because they react faster and with higher yields compared to secondary or tertiary alkyl halides. This is due to less steric hindrance around the carbon bonded to the halogen atom in primary halides.

Out of the given options:

n-Propyl bromide \((CH_3CH_2CH_2Br)\): This is a primary alkyl halide and a good choice for Williamson synthesis.

Isopropyl bromide \((CH_3CHBrCH_3)\): This is a secondary alkyl halide and is less preferred than n-propyl bromide due to increased steric hindrance around the carbon bonded to the bromine atom.

2-Bromo-2-methylpropane \((CH_3C(Br)(CH_3)_2)\): This is a tertiary alkyl halide and is not suitable for Williamson synthesis because tertiary alkyl halides are highly susceptible to undergoing elimination reactions instead of substitution during the reaction.

tert-pentyl bromide \(((CH_3)_3C-Br)\): Similar to 2-bromo-2-methylpropane, this is a tertiary alkyl halide and not ideal for Williamson synthesis due to a high tendency for elimination reactions.

Therefore, the preferred alkyl halide for Williamson synthesis among the given options is: n-Propyl bromide \((CH_3CH_2CH_2Br)\)