Chemistry-Griffin

Diazomethane, CH2N2, is used in the organic chemistry laboratory despite its danger because it produces very high yields and is

dexar [7]

Answer:

See picture below

Explanation:

You are not providing the starting compound. However, I found a similar question so, I will draw the product of this compound, and then, you follow the same procedure.

As the problem states, the diazomethane is often used in reaction with carboxilic acid. This kind of reaction will do an esterification reaction, and the final product will always result in an esther.

So, with the example I give you here, all you have to do is replace the OH in the carboxilic group of your initial compound, for the OCH₃ group.

See picture below.

Hope this helps

5 0

3 years ago

All atoms that react

gregori [183]

Need more to the question

5 0

4 years ago

Question 9(Multiple Choice Worth 1 points)

sergiy2304 [10]

Main sequence starts maybe

6 0

4 years ago

Read 2 more answers

Question 19-20<br> I’m not really sure, can someone help me

Nesterboy [21]

Hi my name is Jaleesa The answer = -1

6 0

3 years ago

Complete this nuclear reaction by selecting which particle would go in the

kompoz [17]

Answer:

Missing particles: $^{0}_{1}e^{+}$ (a positron) and an electron neutrino $\nu_{\rm e}$ .

The nuclear equation would be:

${\rm ^{19}_{10} Ne} \to ^{0}_{1}e^{+} + {\rm ^{19}_{\phantom{1}9}F }+ \nu_{e}$ .

Explanation:

The mass number of a particle is the number on the top-right corner of its symbol.

The atomic number of a particle is the number on the lower-right corner of its symbol.

The nuclear reaction here resembles a beta-plus decay. The mass numbers of the two nuclei are equal. However, the atomic number of the product nucleus is lower than that of the reactant nucleus by $1$ .

A beta decay may either be a beta-plus decay or a beta-minus decay. In a beta-plus decay, a positively-charged positron $^{0}_{1}e^{+}$ and an electron neutrino $\nu_e$ would be released. On the other hand, in a beta-minus decay, a negatively-charged electron $\rm ^{0}_{1}e^{-}$ and an electron antineutrino $\overline{\nu}_e$ would be released.

Electric charge needs to be conserved in nuclear reactions, including this one.

The atomic number of the $\rm Ne$ nucleus on the left-hand side is $10$ , meaning that the nucleus has a charge of $+10$ . On the other hand, the atomic number of the $\rm F$ nucleus on the right-hand side shows that this nucleus carries a charge of only $+9$ .

By the conservation of electric charge, the particles on the right-hand side must carry a positive charge of $+1$ . That rules out the possibility of the combination of one negatively-charged electron $\rm ^{0}_{1}e^{-}$ (with a charge of $-1$ ) and an electron antineutrino $\overline{\nu}_e$ (with no electric charge at all.)

Hence, the only possibility is that the missing particle is a positron (and an electron neutrino $\nu_e$ , which carries no electric charge.)

4 0

3 years ago