Reaction 1
............................O
...........................//
CH2=CH-CH-OH + CH3-CH-CH2-C
........|...........|......\
.......CH3.........CH3......OH
Left of the "+" is COMPOUND 1 and the right is COMPOUND 2.
a) Identify the Functional groups in compounds 1 and 2
b) Complete the equation for reaction 1, by drawing abbreviated structural formula(e) of the product(s) of this reaction.
i) Identify any new functional groups in the organic product(s)
c) State the type of reaction that has occurred and explain how you have arrived at this answer.
If you will look at the post you can see that it is difficult to make sense of it. Probably, from previous posts, I suspect you have OH (ROH are alcohols) and COOH (RCOOH are acids).
RCOOH + R'OH produce esters. Look in your text, notes, and read about esters. The ester produced is as follows:
RCOOH + R'OH ==> RCOOR' + H2O.
........................................................O
......................................................//
CH2=CH-CH-OH + CH3-CH-CH2-C
..............|......................|..............\
............CH3..................CH3............OH
In my question booklet the first CH3 is Under the =CH-"#CH#"-OH and the second CH3 is under CH3-"#CH#"CH2. OH on the right comes off the ...CH2-"C" and so does O.
can you please tell me where you get the R from please and is the ester the reaction that has occured
The R is anything attached to the OH or COOH so ROH is ANY aliphatic alcohol and RCOOH is ANY aliphatic acid. And yes, an alcohol + an acid produces an ester + water. Using the R notation, R denotes any aliphatic group and R' denotes a different (but COULD be the same) aliphatic group; therefore,
RCOOH + HOR' ==> RCOOR' + H2O.
The OH from the RCOOH comes off with the H of the alcohol to form water and the -OR' hooks to the C of the RC=O.
1 answer