University:
Amherst College
Course:
CHEM 221 | Organic Chemistry I
Academic year:

2010
Views:

500

Pages:

4
Author:

Brennen Smith

6 Exam 1 Chem 22 Feb 19, 2010 N.ame page (ja' 6r G K tnr 1 (35) 2 (25) total (100) 3 (25) 4 (20) chk Please avoid defiling this space. . 1 (20 points) Draw the following compounds. Show stereochemistry where relevant. No abbreviations, please . (a) weta- isopropoxyphenol 6 7 ( b) 4-methyl-N-benzylpentanamine (c) c/s-3-(p-ethoxyphenyl)-2-propenal 2. (15 points) For each compound below, how many separate signals would you expect to see in a room -temperature proton-decoupled C NMR spectrum, and how many in a H NMR spectrum? Br £ Br. - e * <=4 Iph # 13 C NMR signals: 3s/-S # ' H NMR signals: /B 7 a H H 6 3. (12 points) Methyl benzoate displays aromatic ' H NMR signals at the chemical shifts indicated . Explain briefly , but clearly , why the ortho and para protons appear downfield of the meta . Some key structures and a few words should be sufficient . H0 6 7.98 Hm 6 7.38 Hp 6 7.48 A. 7rfc^ % + ofe. skr -eJ« $ Huutr *- - /u Orjj » 4 . (13 points) (a ) Cyclohexene’s vinylic Hs appear at 6 5.59. Explain why 2-oxene ( below ) has one vinylic H downfield and the other upfield of cyclohexene’s. Draw structures to support your argument . rvv# .< » f\ A'kfc. 6 6.37 d oJs TVriu ^ / « " 6 4.65 75 C I T \ - S’1 ~J tu. IA ) *( &*s*Or v ~ h sk T-JJS S " = ( b ) 2-Oxene’s C C stretch (IR below ) is much more intense than that of of a simple alkene . Why ? Jw AT! _ /Zv> c =c 3800 3400 3000 2600 2200 2000 1800 1600 1400 1200 wavenumber , cm - 1 lOOO . file. CM 5 7! 600 . 5 (10 points) An unknown ester with the formula C5H10O2 has the H NMR spectrum shown. The integral step heights are in mm. Draw the structure. • ; ' i ; ; I : ; f ' ; ; i ; S : : ! ; : < 3 i ! / 1 : : ; : S ; I ^ • ; ! • -*-**J 40 ,* 4 5 B 2 3 1 6. (15 points) An unknown compound with the formula C5H8C12 displays an IR absorption at about 1670 cm 1 and the !H NMR spectrum below . The numbers of Hs are given . Draw the l , ) structure .! (Assume Js are about 7 Hz unless stated otherwise.) a v* Nc -c a IA ^ '^ / 1 -ct / \;^d V 3H (2^ J = 7 Hz , 1 Hz J=1 Hz / / 2H 2H 2.00 5.90 5.80 5.70 5.60 . 5.50 1H -p- TTT 6.0 5.5 5.0 1 I > p T- -T 4.5 V TTTTT 4.0 3.5 * / S > = C( 3.0 TT 2.5 T •T 2.0 TT 1.5 V 1.0 T 0.5 TTT 0.0 1 7. (20 points) An unknown compound with the formula C j 1H15N displays the IR and H NMR spectra below. Addition of a drop of D2O causes the signal at 8 3.4 to vanish (an HOD peak appears elsewhere); no other signals are affected. The numbers next to the NMR signals indicate the numbers of Hs. Expansions of the downfield signals are included , (a ) Draw the structure . T IOO r rt 90 : J ; 1 \ BO •• I T 70 « A 60 '1 s SO 40 « - 30 r ! T ....i ! A I H : I C E : I :• 20 - to 0 4000 a .. .. 1 M 3800 3400 3600 3200 3000 2SOO T I 2400 2600 ! — -i ... I 2200 2000 - \ 1300 — r- .. . ... j . "a .1 A M . c . • 4 i t -- : iJL ; j : ! I 1400 1SOO 1 1200 600 800 1000 AbO WAVENUMBERS C75 3H » J= > T 5.5 T 5.0 3H 2H 6.5 7.0 1H T 8.0 T 7.5 1 T 7.0 l T T 6.5 6.0 1H 1H i ji T 5.5 1H i I 5.0 I 4.5 J Clj A - C» - T 4.0 T 3.5 T 3.0 y>hx\ s H. \ /v/ / T 2.5 > / A T 2.0 I 1.5 T 1.0 T 0.5 £ H in your structure that is responsible for the signal at 8 5.2, and draw a tree diagram to explain its splitting pattern . ( b) Circle the ppm -cH, - (A C*® b T 0.0 34 ^

Exam 1 Chemistry 22 Key 2010

Related Documents

Description

Related Documents

Free up your schedule!

Take 5 seconds to unlock