(a) From the elemental analysis the empirical formula of
Compound M is C
5H
12O.
(b) Presumably the M
+ ion is not very stable nevertheless its
m/
z value of 88 means that the empirical and molecular mass are the same so the molecular formula of
Compound M is C
5H
12O. The peak at
m/
z = 73 shows loss of a –CH
3 group to give the fragment C
4H
9O
+. –CH
2 can then also be lost to leave the fragment C
3H
7O
+ that is responsible for the peak at
m/
z = 59.
(c) From its molecular formula
Compound M could either be an alcohol or an ether. The very broad absorption centred at 3376 cm
-1 shows the presence of an –OH group so
Compound M is an alcohol. The only other peaks that are easy to attribute are the absorptions centred at about 3000 cm
-1 which are due to C–H.
(d) In the
1H NMR spectrum the integration trace shows that the twelve hydrogen atoms are in four different chemical environments in the ratio of 1:2:6:3. The singlet at 2.8 ppm accounts for the hydrogen atom of the alcohol group. The large singlet due to six protons centred at 1.2 ppm suggests two –CH
3 groups attached to the same carbon atom that has no hydrogen atoms attached to it. The quartet at 1.5 ppm and the triplet at 0.9 ppm represent the –CH
2– and the –CH
3 constituents of an ethyl group respectively. Because the quartet due to the –CH
2– group is not split further this means that the ethyl group is also attached to a carbon atom to which no other hydrogen atoms are attached.
All this information taken together confirms that
Compound M is 2-methylbutan-2-ol, CH
3CH
2C(CH
3)
2OH.