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(e) ? and aqueous sodium hydroxide to give C$_4$H$_4$Na$_2$O$_4$ (f) ? and aqueous ammonia to give C$_4$H$_{10}$N$_2$O$_3$ (g) Methyl benzoate and excess phenylmagnesium bromide, then H$_3$O$^+$ (h) Axetic anhydride and 3-pentanol (i) Ethyl phenylacetate and lithium aluminum hydride, then H$_3$O$^+$ (j) ? and aqueous sodium hydroxide to give C$_4$H$_7$NaO$_3$ (k) ? and aqueous ammonia (l) ? and lithium aluminum hydride, then H$_2$O

          (e)
?
and aqueous sodium hydroxide to give C$_4$H$_4$Na$_2$O$_4$
(f)
?
and aqueous ammonia to give C$_4$H$_{10}$N$_2$O$_3$
(g) Methyl benzoate and excess phenylmagnesium bromide, then H$_3$O$^+$ 
(h) Axetic anhydride and 3-pentanol
(i) Ethyl phenylacetate and lithium aluminum hydride, then H$_3$O$^+$ 
(j)
?
and aqueous sodium hydroxide to give C$_4$H$_7$NaO$_3$
(k)
?
and aqueous ammonia
(l)
?
and lithium aluminum hydride, then H$_2$O
        
Show more…
(e)
?
and aqueous sodium hydroxide to give C4H4Na2O4
(f)
?
and aqueous ammonia to give C4H10N2O3
(g) Methyl benzoate and excess phenylmagnesium bromide, then H3O^+ 
(h) Axetic anhydride and 3-pentanol
(i) Ethyl phenylacetate and lithium aluminum hydride, then H3O^+ 
(j)
?
and aqueous sodium hydroxide to give C4H7NaO3
(k)
?
and aqueous ammonia
(l)
?
and lithium aluminum hydride, then H2O

Added by Teresa H.

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Chemistry: Structure and Properties
Chemistry: Structure and Properties
Nivaldo Tro 2nd Edition
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Could you do EL and please draw the structure? And aqueous sodium hydroxide to give CHNaO. And aqueous ammonia to give C4H10NO. Methyl benzoate and excess phenylmagnesium bromide, then HO. Acetic anhydride and 3-pentanol. Ethyl phenylacetate and lithium aluminum hydride, then HO. And aqueous sodium hydroxide to give CHNaO. CK. And aqueous ammonia. And lithium aluminum hydride, then HO.
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Transcript

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00:03 Question it is given that we need to draw an esterification reaction which is between the propanoic acid and one butanol and we need to show the final product and name that here you'll know that the alcohol react with the acid in the presence of acid catalyst for giving the ester product.
00:50 That is here there is nh2 and 1 2 here nh2 and then double bone door o h plus this is light v and here oh 1 butanol in the presence of h plus it will be giving double bondor and this is nh2 and this is nh2.
01:30 So this is ester product.
01:37 Now second part of the question is that below the structure of lysine that is left and the serene which is right, we need to find the final product when two amino acids forming peptide bonds between them.
02:01 That is the lysine is this is lysine.
02:16 H2 oh oh double bond of.
02:23 So here the amino acid amine group reacting with the another amino acid that is carboxyl group giving peptide bond will be formed.
02:41 Now this peptide bond formed between the amino acid and carboxylic group this is the h2o molecule...
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