CH3CH2COOH splitting pattern q
I know that when looking for a splitting pattern you need to know the number of hydrogens on the 'neighbouring' carbon atom. Is that the carbon on the right hand side, or on the left hand side? As each side produces a different pattern.
Like if the question was, what splitting pattern does the CH2 in CH3CH2COOH produce, would I be looking at the number of H's on the neighbouring C atom on the LHS or RHS?
Like if the question was, what splitting pattern does the CH2 in CH3CH2COOH produce, would I be looking at the number of H's on the neighbouring C atom on the LHS or RHS?
Original post by Magenta96
I know that when looking for a splitting pattern you need to know the number of hydrogens on the 'neighbouring' carbon atom. Is that the carbon on the right hand side, or on the left hand side? As each side produces a different pattern.
Like if the question was, what splitting pattern does the CH2 in CH3CH2COOH produce, would I be looking at the number of H's on the neighbouring C atom on the LHS or RHS?
Like if the question was, what splitting pattern does the CH2 in CH3CH2COOH produce, would I be looking at the number of H's on the neighbouring C atom on the LHS or RHS?
In this case, how many hydrogens are there to the left, and how many are there to the right?
Original post by Nirgilis
In this case, how many hydrogens are there to the left, and how many are there to the right?
well on the LHS of the CH2, the neighbouring carbon atom has 3 hydrogen's, and on the RHS of the CH2, the neighbouring carbon atom has 1 hydrogen, right?
Original post by Magenta96
well on the LHS of the CH2, the neighbouring carbon atom has 3 hydrogen's, and on the RHS of the CH2, the neighbouring carbon atom has 1 hydrogen, right?
If you have a pen and paper, I would try and draw the molecule out and see
(or google it, that works too)Original post by Nirgilis
If you have a pen and paper, I would try and draw the molecule out and see (or google it, that works too)
(or google it, that works too)oh I drew it out, and see the LHS carbon has 3 hydrogen's and the RHS carbon has zero hydrogen's
Original post by Magenta96
oh I drew it out, and see the LHS carbon has 3 hydrogen's and the RHS carbon has zero hydrogen's
Yep
. So using the n+1 rules, you will get 4 peaks. IF you had 3 on the left and 1 on the right, you add them together make 4 hydrogens in total adjacent, then put THAT into n+1 to make 5 peaks. You only really have to worry about weird splittings at University I think 
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