TBR CBT 5 #112 mistake?

[zzomg]

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I've pasted the question below:

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112. The major product for the aldol reaction is:

A. Product I, because all of the peaks are singlets.
B. Product I, because the methyl group adjacent to the carbonyl carbon appears at 5.26 ppm.
C. Product II, because all of the peaks are singlets. C is the best answer. Figure 1 shows the 1H NMR spectrum for the major product of the reaction, so we must determine whether the spectrum fits Product I or Product II. Product I shows both a triplet and quartet in its 1H NMR spectrum because of the ethyl group, with the signal for the CH3 being found near 1.00 ppm. This does not fit the spectrum in Figure 1, which eliminates both choice A and choice B. Because there are four types of hydrogens in Product II, there can be only four signals in its 1H NMR spectrum, so choice D is eliminated. In Product II, all of the hydrogens are isolated on carbons without any neighboring hydrogens. This ensures that all of its peaks are singlets, as seen in Figure 1. The best answer is C.
D. Product II, because it shows a total of five signals in its 1H NMR spectrum.


I see that compound I is a better answer than compound II, but am I missing something or is this compound supposed to have 6 peaks (3 different benzene H's)? Exam this Saturday, want to get this cleared up, would appreciate any insight you guys might have. Thanks!

 

[kehlsh]

Product II is correct because for every carbon adjacent to the carbon that has H to it, there is no hydrogen: thus, every peak has to be a singlet.

5 from "5H" tells you the integration number
meaning, you see 5 hydrogens from benzene, 1 1H and 2 3Hs like the product 2 has.

Lastly, you posted in the incorrect forum.

 

[zzomg]

Product II is correct because for every carbon adjacent to the carbon that has H to it, there is no hydrogen: thus, every peak has to be a singlet.

5 from "5H" tells you the integration number
meaning, you see 5 hydrogens from benzene, 1 1H and 2 3Hs like the product 2 has.

Lastly, you posted in the incorrect forum.

Oh ya I meant I see why II is a better answer than I, if we accept that all the H's on the benzene ring are equivalent. But aren't the ortho, meta, and para hydrogens all different? Like the ortho H's are split by the metas, the metas are split by both the orthos and the para, and the para split by the metas? Or is my understanding of the H NMR splitting wrong?

And whoops you're right about the forum, my bad. I'll move this over.