Consider the above sequence of reactions. 151 g of 2-bromopentane is made to react. Yield of major product P is 80% whereas Q is 100%.

Mass of product Q obtained is _________ g.
(Given molar mass in g mol$$^{-1}$$ H: 1, C: 12, O: 16, Br: 80)

The problem can be solved by first identifying the products formed in each reaction step and then carrying out the required stoichiometric calculations.

In the first step, the starting compound is 2-bromopentane. When treated with alcoholic KOH, it undergoes a $\beta$-elimination reaction.

According to Zaitsev's rule, the more substituted alkene is formed as the major product.

The reaction is

$$\text{2-Bromopentane}+\text{alc. KOH}\longrightarrow\text{Pent-2-ene}+KBr+H_2O.$$

Hence, product P is

$$\boxed{\text{Pent-2-ene }(\mathrm{CH_3-CH=CH-CH_2-CH_3}).}$$

In the second step, pent-2-ene reacts with bromine through electrophilic addition across the double bond.

The reaction is

$$\text{Pent-2-ene}+Br_2\longrightarrow\text{2,3-Dibromopentane}.$$

Therefore, product Q is

$$\boxed{\mathrm{CH_3-CH(Br)-CH(Br)-CH_2-CH_3}.}$$

Next, calculate the molar masses.

For 2-bromopentane,

$$\mathrm{C_5H_{11}Br},$$

the molar mass is

$$M=(5\times12)+(11\times1)+80$$

$$=60+11+80$$

$$=151\ \text{g mol}^{-1}.$$

For 2,3-dibromopentane,

$$\mathrm{C_5H_{10}Br_2},$$

the molar mass is

$$M=(5\times12)+(10\times1)+(2\times80)$$

$$=60+10+160$$

$$=230\ \text{g mol}^{-1}.$$

Initially,

$$151\ \text{g}$$

of 2-bromopentane is taken.

Hence,

$$\text{Number of moles}=\frac{151}{151}=1\ \text{mol}.$$

The first reaction proceeds with an

$$80%$$

yield.

Therefore, the actual amount of product P formed is

$$1\times0.80=0.8\ \text{mol}.$$

The second reaction has

$$100%$$

yield and follows a

$$1:1$$

stoichiometric ratio.

Hence,

$$\text{Moles of Q}=0.8\ \text{mol}.$$

Finally, the mass of product Q is

$$\text{Mass}=0.8\times230$$

$$=184\ \text{g}.$$

Therefore, the required mass of product Q is

$$\boxed{184\ \text{g}}.$$