Reduction of nitroalkanes in neutral medium Zn - NH 4 Cl forms mainlyA- R-N H2B- All of theseC- R-N-N-C lD- R- NHOH

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Synthesis by Reduction of Nitro Compounds

Reduction of ...

Question

Reduction of nitroalkanes in neutral medium ( $Z n / N H_{4} C l$ ) forms mainly

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

All of these

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

Suggest Corrections

Similar questions

Reduction of nitroalkanes in neutral medium ( $Z n / N H_{4} C l$ ) forms mainly

a_{n} = \sum_{r = 0}^{n} \frac{1}{^{n} C_{r}}

\sum_{r = 0}^{n} \frac{r}{^{n} C_{r}}

If $a_{n} = \sum_{r = 0}^{n} \frac{1}{^{n} C_{r}}$ then $\sum_{r = 0}^{n} \frac{r}{^{n} C_{r}}$ equals

N H_{4} C l

^{n} C_{0},^{n} C_{1},^{n} C_{2}, \dots,^{n} C_{n}

(1 + x)^{n}

p + q = 1

n \sum r = 0 r^{2}^{n} C_{r} p^{r} q^{n - r}

Join BYJU'S Learning Program