1. Reaction type involved

The question is from compounds containing nitrogen, so the transformation is the typical exhaustive methylation followed by elimination of an amine, i.e. Hofmann elimination.

In Hofmann elimination, a quaternary ammonium hydroxide on heating undergoes $\beta$-elimination to give an alkene.

A very important rule is:

$\text{Hofmann elimination gives the less substituted alkene as the major product.}$

So the products formed are called Hofmann products, not Saytzeff products.

2. Compare the two possible alkenes

For a pentyl skeleton, the two possible elimination products shown in the options are:

• $\mathrm{CH_3CH_2CH=CHCH_3}$ = pent-2-ene
• $\mathrm{CH_3CH_2CH_2CH=CH_2}$ = pent-1-ene

Now,

• pent-2-ene is the more substituted alkene
• pent-1-ene is the less substituted alkene

Therefore, under Hofmann elimination:

$\mathrm{CH_3CH_2CH_2CH=CH_2}$

is the Hofmann product.

3. Assign A and B

From the given options, the correct statement must identify:

• one product as $\mathrm{CH_3CH_2CH_2CH=CH_2}$
• the other as $\mathrm{CH_3CH_2CH=CHCH_3}$
• and classify them as Hofmann products

Option C states exactly this ordering:

• $A = \mathrm{CH_3CH_2CH_2CH=CH_2}$
• $B = \mathrm{CH_3CH_2CH=CHCH_3}$
• products are Hofmann products

This matches the rule that the less substituted alkene is favored.

4. Check all options

• A: calls them Saytzeff products → incorrect
• B: structures ordered differently and also inconsistent with Hofmann assignment → incorrect
• C: less substituted alkene assigned correctly, classified as Hofmann → correct
• D: same structures as C but calls them Saytzeff products → incorrect

5. Final answer

$\boxed{\text{C}}$

The elimination is Hofmann elimination, giving the less substituted alkene preferentially:

$A = \mathrm{CH_3CH_2CH_2CH=CH_2}, \quad B = \mathrm{CH_3CH_2CH=CHCH_3}$