Lowpass filter impulse response in frequency domain
Discrete time domain comes with its magics and one of the funniest of them out there is the result of the multiplication of an impulse response $h[n]$ by $(-1)^n$, which is equivalent to an alternating version of the original response. 

The key to its understanding comes from a very little effort of converting $(-1)^n$ into a complex exponential and then invoking a fundamental DTFT property, namely: 

if $x[n]$ has the DTFT of $X(e^{j\omega})$ , then $x[n]e^{j\omega_0 n}$ has the DTFT of $X(e^{j(\omega-\omega_0)})$

Therefore, since $(-1)^n = e^{j\pi n}$, then DTFT of  $$h_2[n]=(-1)^n h_1[n]$$ is $$H_2(e^{j\omega}) = H_1(e^{j(\omega - \pi)})$$

Finally the interpretation of $H_1(e^{j(\omega - \pi)})$ is as follows, given that $H_1(e^{j\omega})$ is an ideal lowpass filter of cutoff frequency $\omega_c = \pi/2$ then $H_2(e^{j\omega})$ is a shifted by $\pi$ to the right version of $H_1(e^{j\omega})$ which will be an ideal highpass filter with a cutoff frequecy of $\omega_c = \pi /2$. And the best way to see this will come from drawing the magnitude reponses on the paper...