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Consider the following compound for question 3 & 4. Hon this carbon 3. Answer the following questions about H NMR spectroscop
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Answer #1

1. All protons of each carbon different (non-equivalent) and 3 positions of the benzene ring does not contain proton therefore does not give signals.

2. In 1H-nmr 10 signals would be appeared.

3. -*CH2CH3, 4 mulitiplicity (quartet) are seen.

4. For 13 C- nmr 12 signals would be appeared.

5. And in off- resonance 13C-nmr doublet are seen for -CH- in benzene ring.

6. First structure will be correct.

7 2 orthe HC eneh Carban ditperent Lyically All potons hen-eguirale to Sderhtuled bengene 3 a) Ih H-Nmk appeas Fь) peak atpearas a frerer hatet One to 4 9 2 nmr Sipeals atferar in 13nr 12 bD nmr ar s numba eson4n e 7 pae ton drectly linkt eloublet GM aptet hipler E 2H mulhplur 2 H triplur 4.0 PP 3 H i)-0CH deshield birlel 2H 2 H mnltiplut H3-en en0- 0.98 3H tr0hseve deshielding -o-c4 Sn402 omi ted obs, deeweld tor 2 But tei 2H no Fobsesmed GH nor seen enih

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