Consider all the structural isomers with molecular formula $$C_{5}H_{11}Br$$ are separately treated with KOH(aq) to give respective substitution products, without any rearrangement. The number of products which can exhibit optical isomerism from these is ______.

We need to find all structural isomers of $$C_5H_{11}Br$$, treat each with aqueous KOH (SN2/SN1 substitution, no rearrangement), and count how many products exhibit optical isomerism.

The structural isomers of $$C_5H_{11}Br$$ (monobromopentanes) are:

1. $$CH_3CH_2CH_2CH_2CH_2Br$$ (1-bromopentane)

2. $$CH_3CH_2CH_2CHBrCH_3$$ (2-bromopentane)

3. $$CH_3CH_2CHBrCH_2CH_3$$ (3-bromopentane)

4. $$(CH_3)_2CHCH_2CH_2Br$$ (1-bromo-3-methylbutane)

5. $$(CH_3)_2CHCHBrCH_3$$ (2-bromo-3-methylbutane)

6. $$(CH_3)_2CBrCH_2CH_3$$ (2-bromo-2-methylbutane)

7. $$(CH_3)_3CCH_2Br$$ (1-bromo-2,2-dimethylpropane / neopentyl bromide)

8. $$CH_3CH_2CH(CH_3)CH_2Br$$ (1-bromo-2-methylbutane)

Now, substitution with KOH(aq) replaces Br with OH. The product is the corresponding alcohol. For optical isomerism, the product must have a chiral center (asymmetric carbon).

The products that can exhibit optical isomerism are: 2-pentanol (from isomer 2), 3-methyl-2-butanol (from isomer 5), and 2-methyl-1-butanol (from isomer 8).

The answer is 3.