For peak a its a triplet n1 3 so n 2 theres

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Unformatted text preview: but benzene has 4DUS, and you don't have any left, so that's not it either. Peak A is probably hydrogens next to a carbonyl. A note about integration: these tell the ratio of hydrogens, not the exact number. If the integration adds up to the number of H's, then each number represents the number of hydrogens in each peak. If not then you know the molecule is symmetric: just find out how many hydrogens each integration unit represents. Here, the integrations add to 16, so we're good. The integration of peak A is 2, so it's a CH2. Remember from your splitting rules that you split based on the hydrogens next to you: Ha will be split by Hc. Hb will also be split by Hc. As for Hc, it will be split by both. The multiplicity tells...
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